Solve Smallest Value of Ms for Ladder Equilibrium

[mattmannmf]

So there is a ladder leaning against the wall. its a 2.5 (base), 6 (height), 6.5 (hyp) triangle. the coefficient of static friction is zero at B (corner of 6 & 6.5). we need to determine the smallest value of Ms at A (corner of 2.5 & 6.5) for which equilibrium will be maintained.

They do not give me weight. just dimensions.

I started taking the moment from the middle of the hyp using the Normals at the points as the vectors:

M=0= (NB)(3) - (NA)(1.25)
NA= 3/1.25 NB
where NB is the normal at B and NA is the normal at A

then summed up the forces in the x directions: (FfrA is the friction at point A)
Fx=0= FfrA - NB
NB= Ms NA

then substituted my NA from moments into my sum:

NB= Ms (3/ 1.25 NB)
divided it out and got:
Ms= .4166666

I checked out the answer in the back of the book which is .208, All I need to do is divide the number that I got by 2 and I get .208... any ideas where I went wrong? Thanks

 

[issacnewton]

M=0= (NB)(3) - (NA)(1.25)
NA= 3/1.25 NB
where NB is the normal at B and NA is the normal at A

if you are taking the moment around the center of mass, you forgot to take the moment
of the frictional force at the bottom. N_A is perpendicular to the ground, and at the
moment of slipping, maximum static friction is directed towards the wall.

 

[mattmannmf]

if you are taking the moment around the center of mass, you forgot to take the moment
of the frictional force at the bottom. N_A is perpendicular to the ground, and at the
moment of slipping, maximum static friction is directed towards the wall.

Thanks! got the right answer!